Let $y=\dfrac{x^2+6x}{10-x^2}$. $\dfrac{dy}{dx}=$
$\dfrac{x^2+6x}{10-x^2}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d y d x = d d x ( x 2 + 6 x 10 − x 2 ) = ( 10 − x 2 ) d d x ( x 2 + 6 x ) − ( x 2 + 6 x ) d d x ( 10 − x 2 ) ( 10 − x 2 ) 2 The quotient rule = ( 10 − x 2 ) ( 2 x + 6 ) − ( x 2 + 6 x ) ( − 2 x ) ( 10 − x 2 ) 2 Differentiate ( x 2 + 6 x ) & ( 10 − x 2 ) = 20 x + 60 − 2 x 3 − 6 x 2 + 2 x 3 + 12 x 2 ( 10 − x 2 ) 2 Expand = 6 x 2 + 20 x + 60 ( 10 − x 2 ) 2 \begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2+6x}{10-x^2}\right) \\\\ &=\dfrac{(10-x^2)\dfrac{d}{dx}(x^2+6x)-(x^2+6x)\dfrac{d}{dx}(10-x^2)}{(10-x^2)^2} \gray{\text{The quotient rule}} \\\\ &=\dfrac{(10-x^2)(2x+6)-(x^2+6x)(-2x)}{(10-x^2)^2} \gray{\text{Differentiate }(x^2+6x)\text{ & }(10-x^2)} \\\\ &=\dfrac{20x+60-2x^3-6x^2+2x^3+12x^2}{(10-x^2)^2} \gray{\text{Expand}} \\\\ &=\dfrac{6x^2+20x+60}{(10-x^2)^2} \end{aligned} In conclusion, $\dfrac{dy}{dx}=\dfrac{6x^2+20x+60}{(10-x^2)^2}$, or any other equivalent form.